Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 5;
const int mod = 1e9 + 7;
const double eps = 1e-9;
double sy, a, b;
double fl[maxn], fr[maxn];
double sum[maxn];
int main() {
  scanf("%lf%lf%lf", &sy, &a, &b);
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%lf%lf", &fl[i], &fr[i]);
    sum[i + 1] = sum[i] + fr[i] - fl[i];
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    double x, y;
    scanf("%lf%lf", &x, &y);
    double k = y / (y - sy);
    double ax = (a - x) * k + x;
    double bx = (b - x) * k + x;
    if (bx <= fl[0] || ax >= fr[n - 1]) {
      printf("%lf\n", 0.0);
      continue;
    }
    int l = 0, r = n - 1, m;
    while (l <= r) {
      m = (l + r) / 2;
      if (fr[m] > ax + eps)
        r = m - 1;
      else
        l = m + 1;
    }
    l = r + 1;
    double L = sum[l + 1];
    L -= min(fr[l] - fl[l], fr[l] - ax);
    l = 0, r = n - 1;
    while (l <= r) {
      m = (l + r) / 2;
      if (fl[m] < bx + eps)
        l = m + 1;
      else
        r = m - 1;
    }
    r = l - 1;
    double R = sum[n] - sum[r];
    R -= min(fr[r] - fl[r], bx - fl[r]);
    double ans = (b - a) * (sum[n] - R - L) / (bx - ax);
    printf("%lf\n", ans);
  }
  return 0;
}